Hint for real analysis question

Question

Suppose that $f$ is one-to-one and continuous on [$a,b$]. Prove that $f$ is either strictly increasing or strictly decreasing on [$a,b$].

Suppose it weren't. Use the intermediate value theorem to derive a contradiction. — Daniel Fischer, Oct 23 '13 at 22:30

AlexR · Accepted Answer · 2013-10-23T22:32:08.313

0

Hint Use the intermediate-value theorem to get a contradiction if you assume $$\exists x,y \in [a,b] : f'(x) < 0 \wedge f'(y) > 0$$ Use Weierstraß to acqurie a differentiable function with norm arbitrarily close to $f$.

edited Oct 23 '13 at 22:32

answered Oct 23 '13 at 22:25

AlexR

24,905

Most (in the sense that the complement is of the first category) elements of $C^0([a,b])$ are nowhere differentiable. – Daniel Fischer Oct 23 '13 at 22:27
@All sorry, updated argument. Weierstraß works here. – AlexR Oct 23 '13 at 22:32

score 0 · Answer 2 · answered Oct 23 '13 at 22:32

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Hint: Go by contradiction. If its not monotone then we have some $p \le q \le r$ such that

$$f(p) \ge f(q) \le f(r)$$

Then apply the intermediate value theorem to get a contradiction.

answered Oct 23 '13 at 22:32

Deven Ware

7,076

Hint for real analysis question

2 Answers2